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livestrong136 livestrong136
wrote...
Posts: 324
Rep: 7 0
12 years ago
MY MISTAKE IS: correct that the critical points are -2 and 3, but you did not determine the nature of these values i.e. what is going on at these values (-2 marks)

1.Given the function M(t) = 2t3 – 3t2 – 36t, find the critical values and determine, using both the second derivative test and a sign chart, the nature of these values.

M(t) = 2t3 – 3t2 – 36t             M’(t) = 6t2 – 6t – 36
M’(t) = 6t2 – 6t – 36             M”(t) = 12t – 6
0 = 6t2 – 6t – 36             t = 0.5             
t = -2, 3

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BearProBearPro
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12 years ago
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